Questions: Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the y-axis. x=-y^2+5y

Solution Steps

The region to be revolved is bounded by the curve \( x = -y^2 + 5y \) and the y-axis.

The limits of integration are determined by the points where the curve intersects the y-axis. Setting \( x = 0 \) in the equation \( x = -y^2 + 5y \), we solve for \( y \): \[ 0 = -y^2 + 5y \] \[ y(y - 5) = 0 \] Thus, \( y = 0 \) and \( y = 5 \). Therefore, the limits of integration are from \( y = 0 \) to \( y = 5 \).

The volume \( V \) of the solid formed by revolving the region about the y-axis can be found using the disk method. The radius of each disk is given by the x-coordinate of the curve, \( x = -y^2 + 5y \). The volume is given by: \[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy \] where \( R(y) = -y^2 + 5y \) and the limits of integration are from \( y = 0 \) to \( y = 5 \).

\[ V = \pi \int_{0}^{5} (-y^2 + 5y)^2 \, dy \]

\[ (-y^2 + 5y)^2 = y^4 - 10y^3 + 25y^2 \]

\[ V = \pi \int_{0}^{5} (y^4 - 10y^3 + 25y^2) \, dy \] \[ V = \pi \left[ \frac{y^5}{5} - \frac{10y^4}{4} + \frac{25y^3}{3} \right]_{0}^{5} \] \[ V = \pi \left[ \frac{5^5}{5} - \frac{10 \cdot 5^4}{4} + \frac{25 \cdot 5^3}{3} \right] \]

\[ V = \pi \left[ 625 - \frac{10 \cdot 625}{4} + \frac{25 \cdot 125}{3} \right] \] \[ V = \pi \left[ 625 - 1562.5 + 1041.67 \right] \] \[ V = \pi \left[ 625 - 1562.5 + 1041.67 \right] \] \[ V = \pi \left[ 104.17 \right] \]

Final Answer